GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 25Crafting an Impactful Resume: Balancing Detail with Brevity
10:00 AM EDT
-10:30 AM EDT
Showcase your achievements through impactful storytelling! Get ready for an LIVE session on April 25, Learn how to craft compelling narratives on your resume that resonate with admission committees.
Apr 24Get an Extra 30% Off Target Test Prep Plans
08:00 PM EDT
-11:59 PM EDT
Grab 30% off any Target Test Prep GRE plan during our Flash Sale and join Brian and many other students who have used Target Test Prep to score high on the GRE. Just enter the coupon code FLASH30 at checkout to save big.
Apr 26Get a Top GRE Score with Target Test Prep
12:00 PM EDT
-01:00 PM EDT
The Target Test Prep course represents a quantum leap forward in GRE preparation, a radical reinterpretation of the way that students should study. In fact, more and more Target Test Prep students are scoring 329 and above on the GRE.
The probability of rain is on any given day next week.
[#permalink]
05 Aug 2018, 14:44
05 Aug 2018, 14:44
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
65% (00:58) correct
34% (00:59) wrong based on 150 sessions
Hide Show timer Statistics
The probability of rain is \(\frac{1}{2}\) on any given day next week.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability that it rains on at least one of the 7 days next week |
\(\frac{127}{128}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The probability of rain is on any given day next week.
[#permalink]
07 Aug 2018, 05:09
07 Aug 2018, 05:09
2
The probability that it rains on one of 7 days = 1-probability that it does not rain on any of the 7 days.
probability for not raining on any of the 7 days = \((1/2)^7 = \frac{1}{2^7}\)
ans = \(1 - \frac{1}{2^7}\)
or, \(\frac{(2^7 - 1)}{2^7} = \frac{127}{128}\)
probability for not raining on any of the 7 days = \((1/2)^7 = \frac{1}{2^7}\)
ans = \(1 - \frac{1}{2^7}\)
or, \(\frac{(2^7 - 1)}{2^7} = \frac{127}{128}\)
Re: The probability of rain is on any given day next week.
[#permalink]
09 Aug 2018, 15:09
09 Aug 2018, 15:09
2
Expert Reply
Explanation
Since Quantity A is an “at least” problem, use the 1 – x shortcut. Rather than calculate the probability of rain on exactly 1 day next week, and then the probability of rain on exactly 2 days next week, and so on (after which you would still have to add all of the probabilities together!), instead calculate the probability of no rain at all on any day, and then subtract that number from 1.
That will give the combined probabilities for any scenarios that include rain on at least 1 day. Probability of NO rain for any of the 7 days =\(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{128}\)
Subtract this probability from 1: \(1-\frac{1}{128}=\frac{127}{128}\).
Quantities A and B are equal.
Since Quantity A is an “at least” problem, use the 1 – x shortcut. Rather than calculate the probability of rain on exactly 1 day next week, and then the probability of rain on exactly 2 days next week, and so on (after which you would still have to add all of the probabilities together!), instead calculate the probability of no rain at all on any day, and then subtract that number from 1.
That will give the combined probabilities for any scenarios that include rain on at least 1 day. Probability of NO rain for any of the 7 days =\(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{128}\)
Subtract this probability from 1: \(1-\frac{1}{128}=\frac{127}{128}\).
Quantities A and B are equal.
